Partial derivative of an integral of several variables

**Yair1978** · July 14th 2011, 09:24 PM

Hi,

I'm trying to understand Fourier series and in order to do so have to be able to calculate the partial derivates of the function:
I_n=(Integral [ f(x) - a_0/2 - sigma (a_k cos(kx) + b_k sin(kx)]^2)dx
w.r.t. a_0, a_k and b_k, where the integral is calculated from -pi to pi.

I actually had certain thoughts about how to go about this, which I originally wrote here, but as I quickly realized it was completely wrong I've deleted this... Now I'm thinking like this: it's a definite integral, so if I calculate the primitive functions of all the terms containing x, and then calculate the partial derivatives that should do it, no?

Still, I don't get the answer in my book... So....

If someone could explain to me how to go about this in an easy-to-understand, step-by-step manner it'd be greatly appreciated!

/Yair

**HallsofIvy** · July 16th 2011, 05:21 AM

This is $\int_{-\pi}^{\pi} [f(x)- \frac{a_0}{2}- \sum (a_k cos(kx)+ b_k sin(kx)]^2 dx$ ?

The partial derivatives of $\int_a^b f(x, u, v)dx$ with respect to u and v are
$\int_a^b \frac{\partial f}{\partial u} dx$ and $\int_a^b \frac{\partial f}{\partial v}dx$ .

That is, just take the partial derivatives of the integrand.

**Yair1978** · July 16th 2011, 11:57 AM

Originally Posted by HallsofIvy

This is $\int_{-\pi}^{\pi} [f(x)- \frac{a_0}{2}- \sum (a_k cos(kx)+ b_k sin(kx)]^2 dx$ ?

Yes, that's what I meant.

Originally Posted by HallsofIvy

The partial derivatives of $\int_a^b f(x, u, v)dx$ with respect to u and v are
$\int_a^b \frac{\partial f}{\partial u} dx$ and $\int_a^b \frac{\partial f}{\partial v}dx$ .

That is, just take the partial derivatives of the integrand.

I see what you're saying. It'd be great if you could prove this for me!

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